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Domain and range of this function? 
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#1
May2108, 01:20 AM

P: 18

1. The problem statement, all variables and given/known data
(1/(x+7))5 find the domain and range. im having trouble graphing this one on my calculator, because when i zoom out the graph looks a lot different than when i zoom in. i think i have an answer though, so i need a quick check. domain: none (infinite) rangE: 0 to 10? 


#2
May2108, 01:22 AM

P: 1,754

Try with no calculator.
Can the denominator equal 0? Is a function defined with a 0 denominator? Set your denominator equal to 0, solve for x. Range: R 


#3
May2108, 01:24 AM

P: 18

so how do i calculate the range without a calculator? 


#4
May2108, 01:37 AM

P: 18

Domain and range of this function?
BUMP! GAH i have 24 minutes to complete this :(



#5
May2108, 02:00 AM

HW Helper
P: 2,616

Note, as rocomath said, the denominator, and hence the function, is undefined when it is zero. What does that tell you about the possible values of x, ie. the domain?
You don't need a calculator to figure out the range. Let [tex]y = \frac{1}{x+7}  5[/tex]. We want to know the possible values of y. Start by expressing x in terms of y. Once you have done that, look the resulting expression. What values of y are not allowed? The range of the function would then be R\{y} (all real numbers excluding those values of y which is not allowed). 


#6
May2308, 06:28 PM

P: 225

If you really care, the whole point is you can't divide by zero....so we do not want
[itex] \frac{1}{x+7} = \frac{1}{0}[/itex]. That's what he meant by setting the bottom equal to zero, since we see that [itex] x + 7 = 0 [/itex] when x = 7. Thus our domain is all real numbers not including 7. 


#7
May2408, 12:09 PM

Math
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Thanks
PF Gold
P: 39,682

To find the range, reverse the function. Solve for x as a function of y. Now, what values can y have that won't give a "division by 0"?
(What you are reallying doing is finding the inverse function that reverses "domain" and "range".) 


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